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awk to calculate date and show data

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# 1  
Old 11-20-2014
awk to calculate date and show data
data:
Code:
hostcomment {
host_name=myhost01
entry_type=1
comment_id=1
source=0
persistent=1
entry_time=1415723753
expires=0
expire_time=0
author=hpsm
comment_data=IM0837437472
}

program {
modified_host_attributes=1
modified_service_attributes=1
enable_notifications=1
active_service_checks_enabled=1
passive_service_checks_enabled=1
active_host_checks_enabled=1
passive_host_checks_enabled=1
enable_event_handlers=1
obsess_over_services=0
obsess_over_hosts=0
check_service_freshness=1
check_host_freshness=0
enable_flap_detection=1
enable_failure_prediction=1
process_performance_data=0
global_host_event_handler=
global_service_event_handler=
next_comment_id=177361
next_downtime_id=1039
next_event_id=418881
next_problem_id=194777
next_notification_id=59919
}

hostcomment {
host_name=myhost01
entry_type=1
comment_id=1
source=0
persistent=1
entry_time=1415723753
expires=0
expire_time=0
author=hpsm
comment_data=IM023434343
}

i have a file that contains chunks similar to the above.

i want to parse this file and skip over chunks that have entry time which is older than 60 days. Meaning, DO NOT output those chunks.

Please note, some of the chunks in the data file do not have "entry_time". so for those entries, i want this script to output the chunk and move on to the next chunk.

So in other words, i want to do the date subtraction only on chunks that have "entry_time".


this is the code i'm using:
Code:
NOW=$(date +%s)

FILE=${1}

awk -v NOWTIME=$NOW -v mac="=" '
    BEGIN { 
        RS = "{"; 
        FS = " "; 
    } 
    FNR == 1 {
        record_sep = RT;
    }
    { 

        for (i = 1; i <= NF; i++ ) { 
            if ( match( $i, mac ) > 0 ) {
                print record_sep , $0;
                break;
            }
        } 
    }
' ${FILE}

how do i substract the entry time from the time now??? and also, is this script good enough to do what i need?
Last edited by SkySmart; 11-20-2014 at 07:34 PM..
# 2  
Old 11-20-2014
IMO you are better off using perl for this problem...the "entry_time" is given in seconds since the *nix Epoch...and if your awk can figure out the current Epoch then you'd be able to subtract the "now" Epoch from the "entry_time" Epoch and dole out the desired chunks to an output file.

perl comes with the date and time routines builtin...so if I were you I'd doing this in perl otherwise feel free to ignore this post...
# 3  
Old 11-20-2014
Quote:
Originally Posted by shamrock
IMO you are better off using perl for this problem...the "entry_time" is given in seconds since the *nix Epoch...and if your awk can figure out the current Epoch then you'd be able to subtract the "now" Epoch from the "entry_time" Epoch and dole out the desired chunks to an output file.

perl comes with the date and time routines builtin...so if I were you I'd doing this in perl otherwise feel free to ignore this post...
unfortunately, i cant write in a language that i'm unfamiliar with. if this can be done in perl, please, if you can, supply the perl code.

im pretty sure theres goto be a way around this with awk.
# 4  
Old 11-20-2014
That'd depend on what system you are on ie your OS...because if you are using gawk then you certainly can...just look up the systime function inside the gawk man page...
# 5  
Old 11-20-2014
its a linux red hat host.
# 6  
Old 11-20-2014
Quote:
Originally Posted by SkySmart
its a linux red hat host.
Then you probably have gawk so "man gawk" and search for "systime"...
# 7  
Old 11-20-2014
Quote:
Originally Posted by shamrock
if your awk can figure out the current Epoch then you'd be able to subtract the "now" Epoch from the "entry_time" Epoch and dole out the desired chunks to an output file.
Although it is not a well known fact, all awk releases are able to retrieve the current epoch time by standard :
Code:
$ awk 'BEGIN {srand();time=srand();print time}'
1416532245

These 2 Users Gave Thanks to jlliagre For This Post:
shamrock vgersh99
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